Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n=sc.nextInt();
    double[] a=new double[n];
    for(int i=0;i<n;i++){
      a[i]=sc.nextDouble();
    }
    double[][] dp = new double[n+2][n+2];
    dp[0][0] = 1-a[0];
    dp[0][1] = a[0];
    for(int i=0;i<n-1;i++){
      for(int j=0;j<=i+1;j++){
        dp[i+1][j]+=dp[i][j]*(1-a[i+1]);
        dp[i+1][j+1]+=dp[i][j]*a[i+1];
      }
    }
    double total=0;
    for(int i=n/2+1;i<=n;i++){
      total+=dp[n-1][i];
    }
    System.out.println(total);
  }
}